2015 AMC 10A Problems/Problem 2


A box contains a collection of triangular and square tiles. There are $25$ tiles in the box, containing $84$ edges total. How many square tiles are there in the box?

$\textbf{(A)}\ 3\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 7\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 11$


Let $a$ be the amount of triangular tiles and $b$ be the amount of square tiles.

Triangles have $3$ edges and squares have $4$ edges, so we have a system of equations.

We have $a + b$ tiles total, so $a + b = 25$.

We have $3a + 4b$ edges total, so $3a + 4b = 84$.

Multiplying the first equation by $3$ on both sides gives $3a + 3b = 3(25) = 75$.

Second equation minus the first equation gives $b = 9$, so the answer is $\boxed{\textbf{(D) }9}$.

Solution 2

If all of the tiles were triangles, there would be $75$ edges. This is not enough, so there needs to be some squares. Trading a triangle for a square results in one additional edge each time, so we must trade out $9$ triangles for squares. Answer: $\boxed{\textbf{(D) }9}$

Solution 3

Let $x$ be the number of square tiles. A square has $4$ edges, so the total number of edges from the square tiles is $4x$. There are $25$ total tiles, which means that there are $25-x$ triangle tiles. A triangle has $3$ edges, so the total number of edges from the triangle tiles is $3(25-x)$. Together, the total number of edges is $4x+3(25-x)=84$. Solving our equation, we get that $x=9$ which means that our answer is $\boxed{\textbf{(D) }9}$.



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See Also

2015 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 10 Problems and Solutions

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